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Question Number 204621 by hardmath last updated on 23/Feb/24

$$\mathrm{a},\mathrm{b},\mathrm{c}\in {\mathbb{R}}^{+}$$$$\mathrm{If}\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}=1$$$$\mathrm{Prove}\mathrm{that}:\mathrm{a}+\mathrm{b}+\mathrm{c}⩾\frac{1}{3}$$

Answered by A5T last updated on 23/Feb/24

$$\frac{a+b+c}{3}⩾{\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)}^2=\frac{1}{9}\Rightarrow a+b+c⩾\frac{1}{3}$$

Answered by mr W last updated on 23/Feb/24

$$a+b+c=\frac{{\left(\sqrt{a}\right)}^2}{1}+\frac{{\left(\sqrt{b}\right)}^2}{1}+\frac{{\left(\sqrt{c}\right)}^2}{1}$$$$⩾\frac{{(\sqrt{a}+\sqrt{b}+\sqrt{c})}^2}{1+1+1}=\frac{1^2}{3}=\frac{1}{3}$$

Commented by mr W last updated on 23/Feb/24

Answered by mr W last updated on 24/Feb/24

Commented by mr W last updated on 24/Feb/24

$$say\sqrt{a}=x,\sqrt{b}=y,\sqrt{c}=z$$$$x+y+z=1istheplane\mathrm{\Delta }ABC.$$$$P(x,y,z)isapointontheplane.$$$$a+b+c=x^2+y^2+z^2={OP}^2$$$${OP}_{min}isthedistancedfromOtothe$$$$plane\mathrm{\Delta }ABC.$$$${OP}_{min}=d=\frac{\mid 0+0+0-1\mid }{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$$$${OP}_{min}^2=\frac{1}{3}$$$$\Rightarrow a+b+c={OP}^2⩾{OP}_{min}^2=\frac{1}{3}$$

Commented by hardmath last updated on 24/Feb/24

$$\mathrm{Perfect}\mathrm{solution},\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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